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Question
Evaluate `int"e"^x (1/x - 1/x^2) "d"x`
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Solution
Let I = `int"e"^x (1/x - 1/x^2) "d"x`
Put f(x) = `1/x`
∴ f'(x) = `-1/x^2`
∴ I = `int"e"^x ["f"(x) + "f'"(x)] "d"x`
= `"e"^x*"f"(x) + "c"`
∴ I = `"e"^x* 1/x + "c"`
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